91Ó°ÊÓ

The mass of proton is $m_p=1836m_e$

The initial speed of neutron is u = 0

The kinetic energy of a particle depends on the mass and speed of the particle. The kinetic energy of particle changes if speed of particle changes and the change in kinetic energy is same as the work done by particle to change the kinetic energy.

Apply the momentum conservation to find the speed of proton:

$Mu=m_p{v}_p-m_e{v}_e$

Here, M is the mass of the neutron.

Substitute all the values in the above equation.

$$\begin{gathered} M\left(0\right)=\left(1836m_e\right)v_p-m_e{v}_e \\ {v}_p=\frac{v_e}{1836} \end{gathered}$$

The total kinetic energy of proton and electron is given as:

$K_T=\frac{1}{2}{m}_p{v}_p^2+\frac{1}{2}{m}_e{v}_e^2$

Substitute all the values in the above equation.

$$\begin{gathered} K_T=\frac{1}{2}\left(1836m_e\right){\left(\frac{v_e}{1836}\right)}^2+\frac{1}{2}{m}_e{v}_e^2 \\ {K}_T=\frac{1}{2}{m}_e\left(\frac{v_e^2}{1836}+v_e^2\right) \\ {K}_T=\frac{1}{2}{m}_e\left(\frac{1837v_e^2}{1836}\right) \end{gathered}$$

The kinetic energy of proton is given as:

$$\begin{gathered} K_p=\frac{1}{2}{m}_p{v}_p^2 \\ {K}_p=\frac{1}{2}\left(1836m_e\right){\left(\frac{v_e}{1836}\right)}^2 \\ {K}_p=\frac{1}{2}{m}_e{\left(\frac{{v^2}_e}{1836}\right)}^2 \end{gathered}$$

The fraction of total kinetic energy released goes into kinetic energy of proton is given as:

$$\begin{gathered} \frac{K_P}{K_T}=\frac{{\displaystyle \frac{1}{2}}{m}_e\left({\displaystyle \frac{v_e^2}{1836}}\right)}{\frac{1}{2}{m}_e\left(\frac{1837v_e^2}{1836}\right)} \\ \frac{K_P}{K_T}=\frac{1}{1837} \end{gathered}$$

Therefore, the fraction of total kinetic energy released goes into kinetic energy of proton is $\frac{1}{1837}$.